astridantonio225
13.04.2021 •
Chemistry
A chemist fills a reaction vessel with 0.247 M lead(II) (Pb2+) aqueous solution, 0.758 M bromide (Br-) aqueous solution, and 0.109 g lead (II) bromide (PbBr2) solid at a temperature of 20.0 oC. Under these conditions, calculate the reaction free energy deltaG for the following chemical reaction: Pb2+(aq) + 2Br-(aq) <> PbBr2(s)
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Ответ:
ΔG = -24.7kJ/mol
Explanation:
ΔG° of
Pb2+(aq) + 2Br-(aq) ⇄ PbBr2(s)
is:
ΔG° PbBr2 - (2*ΔG°Br- + ΔG°Pb2+)
-261.9kJ/mol - (2*-104.0kJ/mol + -24.4kJ/mol) =
-29.5kJ/mol
ΔG of the reaction is:
ΔG = ΔG° + RT lnQ
Where R is gas constant (8.314x10⁻³kJ/molK)
T is absolute temperature (20°C + 273.15 = 293.15K)
Q is reaction quotient = 1 / [Pb²⁺][Br⁻]²
Replacing:
ΔG = -29.5kJ/mol + 8.314x10⁻³kJ/molK*293.15K ln(1 / [Pb²⁺][Br⁻]²)
ΔG = -29.5kJ/mol + 8.314x10⁻³kJ/molK*293.15K ln(1 / [0.247M][0.758M]²)
ΔG = -24.7kJ/molОтвет:
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Explanation:
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